Time taken by an object falling from rest to cover the height of $h_1$ and $h_2$ is respectively $t_1$ and $t_2$,then the ratio of $t_1$ to $t_2$ is:

$A$ body $A$ is projected upwards with a velocity of $98 \, m/s$. $A$ second body $B$ is projected upwards with the same initial velocity but after $4 \, s$. Both the bodies will meet after how many seconds (from the start of body $A$)?

$A$ body at rest falls through a height $h$ with velocity $V$. If it has to fall down further for its velocity to become $3V$,the distance travelled in that interval is: (in $h$)

$A$ particle is projected vertically upwards from a point $A$ on the ground. It takes $t_1$ time to reach a point $B$ but it still continues to move up. If it takes $t_2$ time to reach the ground from point $B$,then the height of point $B$ from the ground is

$A$ player throws a ball upwards with an initial speed of $29.4 \; m s^{-1}$. What is the direction of acceleration during the upward motion of the ball?

$A$ ball is thrown vertically upwards from the ground $G$ with a speed $u$. It reaches a point $B$ at a height $h$ (lower than the maximum height) after time $t_1$. It returns to the ground after time $t_2$ from the instant it was at $B$ during the upward journey. Then $t_1t_2$ is equal to